Question

If the solution of the differential equation \((2x + 3y - 2) \, dx + (4x + 6y - 7) \, dy = 0, \quad y(0) = 3,\) is  \(\alpha x + \beta y + 3 \log_e |2x + 3y - \gamma| = 6,\) then \(\alpha + 2\beta + 3\gamma\) is equal to ____.

Answer 1

Correct Answer: 29

Given the differential equation:

\((2x + 3y - 2)dx + (4x + 6y - 7)dy = 0, \quad y(0) = 3\)
We define:
\(t = 2x + 3y - 2\)
Differentiating with respect to \( x \):

\(\frac{dt}{dx} = 2 + 3 \frac{dy}{dx}\)
 

Rearranging:

\(\frac{dy}{dx} = \frac{\frac{dt}{dx} - 2}{3}\) 

Step 1. Substituting into the Original Equation: Substituting \( \frac{dy}{dx} \) into the given differential equation:

 \((2x + 3y - 2)dx + (4x + 6y - 7) \left( \frac{\frac{dt}{dx} - 2}{3} \right) dx = 0\)

Step 2. Simplifying:
 
 \(3(2x + 3y - 2) + (4x + 6y - 7) \left( \frac{dt}{dx} - 2 \right) = 0\)
  
Further simplification leads to separation of terms and integration.
Integrating Both Sides: Integrating both sides with respect to \( x \) yields:

\(\int ...\) 

Step 3. Solving for Constants: Given the initial condition \( y(0) = 3 \), we can find the value of constants.

Step 4. Finding the Value of \( \alpha, \beta, \gamma \)**: Substituting known values, we find:

\(\alpha + 2\beta + 3\gamma = 29\)